Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{-2r^2 + 14r - 20}{-r^2 + 25}$
First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {-2(r^2 - 7r + 10)} {-1(r^2 - 25)} $ $ k = \dfrac{2}{1} \cdot \dfrac{r^2 - 7r + 10}{r^2 - 25} $ Next factor the numerator and denominator. $ k = 2 \cdot \dfrac{(r - 5)(r - 2)}{(r - 5)(r + 5)}$ Assuming $r \neq 5$ , we can cancel the $r - 5$ $ k = 2 \cdot \dfrac{r - 2}{r + 5}$ Therefore: $ k = \dfrac{ 2(r - 2)}{ r + 5 }$, $r \neq 5$